Problem: Let $f(x)=2x-3\sin(x)$. $f'(x)=$
Solution: The expression for $f(x)$ includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} f'(x)&=\dfrac{d}{dx}[2x-3\sin(x)] \\\\ &=2\dfrac{d}{dx}(x)-3\dfrac{d}{dx}[\sin(x)] \\\\ &=2\cdot1-3\cdot\cos(x) \\\\ &=2-3\cos(x) \end{aligned}$ In conclusion, $f'(x)=2-3\cos(x)$